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In physics and chemistry, there is the concept of double groups. These are double covers of the usual point groups, obtained by "adding an element $R$, which represents rotation by $2\pi$" to the respective point group.

Since the set $\{ E,R \}$ forms a group isomorphic to $\mathbb Z_2$, it seems that the double group should be some kind of a product of the point group and $\mathbb Z_2$. However, it is not the direct product, since $R$ does not commute with all the elements. Can it be defined a different operation between the point group and $\mathbb Z_2$?

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    $\begingroup$ Is anybody able to translate the gibberish of the Wikipedia page into a mathematical definition? $\endgroup$
    – YCor
    Commented Feb 14 at 17:58
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    $\begingroup$ Is it not just a semidirect product? (en.wikipedia.org/wiki/Semidirect_product) $\endgroup$
    – Sam Hopkins
    Commented Feb 14 at 18:01
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    $\begingroup$ @SamHopkins, re, it is an extension by $\mathbb Z/2$, but it need not split. $\endgroup$
    – LSpice
    Commented Feb 14 at 18:32

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As far as I can tell, a double group is a double cover of a group. Specifically, if $G \subset \operatorname{SO}(n)$ is a group acting by rotations of $n$-dimensional space, its double group is the lift $2G \subset \operatorname{Spin}(n)$ defined as the fibre product $$2G = G \times_{\operatorname{SO}(n)} \operatorname{Spin}(n).$$ It seems that this language is only used when $n=3$ and $G$ is finite. (The point group is the symmetries of a lattice that preserve a point.) Probably chemists don't know about higher-dimensional spaces.

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    $\begingroup$ Probably the mathematical content of this answer can do without the funny but unnecessary dig at chemists at the end. $\endgroup$
    – LSpice
    Commented Feb 14 at 18:33
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    $\begingroup$ Since some chemical structures are symmetrical under reflections, I would guess sometimes $G$ may be a subgroup of $\mathrm{O}(n)$ instead of $\mathrm{SO}(n)$, in which case the spin group could be replaced with a pin group. But which one? $\endgroup$
    – Will Sawin
    Commented Feb 14 at 18:34
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    $\begingroup$ Less snarkily put: Chemists are less concerned with such groups in higher dimensions. $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Feb 14 at 23:36
  • $\begingroup$ @WillSawin Good question. I don't know much chemistry, just a little physics, so I can only speculate. My guess is that what you want to know is the possible eigendistributions of electrons, where you know that you have n electrons per site of your lattice. This is basically a question about the representation theory of the point group: the final answer will be a free module for the translation group, so it will be induced up from a representation of the point group (aka little group). If you have an odd number of electrons, then you need to look at spin representations aka reps of the ... $\endgroup$
    – Theo Johnson-Freyd
    Commented Feb 15 at 22:42
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    $\begingroup$ But that's a dismissive response. Here is what I think the actual answer is. The actual answer is probably: "It doesn't matter which of the Pin groups you use". The reason is that what you care about is the block of Rep(2G) in which the centre acts nontrivially. This block is the same for both Pin groups. Indeed, it is the same as a Rep(G)-module category. Indeed, if you arbitrarily choose a $\sqrt{-1}$, then, writing $2_\pm G$ for the $\mathrm{Pin}_\pm$-cover of $G \subset \mathrm{O}(n)$, what you find is that every $2_+ G$-module is also canonically a $2_- G$-module, and conversely. The ... $\endgroup$
    – Theo Johnson-Freyd
    Commented Feb 15 at 22:52

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